during the night only 3 units are demanded. Total output for each 24hour period is 7 units. The utility
produces electricity according to the production function
y
i
= (
KF
i
)
1
/
2
,
i
=
day, night
,
where
K
is the size of the generating plant, and
F
i
is tons of fuel. The °rm must build a single plant; it
cannot change plant size from day to night. If plant size costs
w
k
per 24hour period and a ton of fuel costs
w
f
, what size plant will the utility build?
Fuel needed per day and per night:
F
d
=
y
2
d
K
F
n
=
y
2
n
K
Total costs per day (24 hours) are:
TC
=
w
k
K
+
w
f
(
F
d
+
F
n
)
→
min
K
!
Choose plant size to minimize cost:
TC
=
w
k
K
+
w
f
y
2
d
+
y
2
n
K
→
min
K
!
FOC:
w
k
+
w
f
y
2
d
+
y
2
n
K
2
(

1)
!
= 0
w
f
(
y
2
d
+
y
2
n
)
d
1
K
2
=
w
k
K
=
s
w
f
(
y
2
d
+
y
2
n
)
w
k
Since demand is
y
d
= 4
and
y
n
= 3
,
K
=
r
w
f
(16 + 9)
w
k
=
r
25
w
f
w
k
= 5
r
w
f
w
k
.
13
8.
Suppose that a °rm has production function
Q
= 2
L
√
K
.
Remark:
CobbDouglas production function with
A
= 2
, α
L
= 1
, α
K
=
1
2
.
a)
If the price of labour is 2 and the price of capital is 4, what is the optimal input mix?
Now we have
p
L
= 2
,
p
K
= 4
. Cost minimisation program:
min
K,L
2
·
L
+ 4
·
K
s.t.
(2
L
√
K

Q
) = 0
⇒ L
= 2
L
+ 4
K

λ
(2
L
√
K

Q
)
FOC:
∂
L
∂K
= 4

1
2
·
2
·
L
√
K
λ
!
= 0
⇒
λ
=
4
√
K
L
∂
L
∂L
= 2

2
λ
√
K
!
= 0
⇒
λ
=
1
√
K
⇒
4
√
K
L
=
1
√
K
⇔
4
K
=
L
This last result describes the optimal input mix. Insert this into the constraint:
2
L
√
K

Q
= 0
8
K
√
K

Q
= 0
8
K
3
/
2

Q
= 0
⇒
K
=
3
s
Q
8
2
=
Q
8
2
/
3
⇒
L
= 4
·
Q
8
2
/
3
Summary:
°
slope of isocost line:
p
L
p
K
=
1
2
°
slope of isoquant:

∂Q/∂L
∂Q/∂K
=
2
√
K
L
√
K
=
2
K
L
In the optimum, the isoquant touches the isocost line, i. e. their slopes are equal:
1
2
=
2
K
L
⇔
L
= 4
K
This is the same condition as above.
b)
For an output level of
Q
= 1000
, how much of each input will be used?
see a):
K
=
Q
8
2
/
3
=
1
,
000
8
2
!
1
/
3
=
3
r
1
,
000
,
000
64
=
100
4
= 25
⇒
L
= 4
·
K
= 100
14
c)
How should inputs be scaled to attain
Q
= 8000
? What will be the percentage increase in cost?
f
(
sK, sL
) = 2
·
(
s
·
L
)
√
sK
!
= 8
,
000
⇔
s
3
/
2
(2
L
√
K
)

{z
}
for
Q
=1
,
000
= 8
,
000
⇒
s
3
/
2
= 8
⇒
s
= 8
2
/
3
=
3
√
8
2
= 4
where we know from part (b) that
2
L
√
K
is the input mix that yields output
Q
= 1
,
000
at
minimum cost.
With this production function, quadruple labour and capital use to obtain 8,000 units of out
put. Due to
s
3
/
2
we have increasing returns to scale. They are increasing because
α
L
= 1
and
α
K
=
1
2
, so
α
L
+
α
K
=
3
2
>
1
.
15
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 Winter '16
 Robledo
 Economics, Microeconomics